United States Member #116344 September 8, 2011 3941 Posts Offline

Posted: November 9, 2013, 8:46 pm - IP Logged

This is a tentative formula derived from ' finding the next term' in a sequence. Whether such sequence is linear or not is left to observation. I use the word 'tentative' to avoid any equivocation . I will replace the Nth term, with Nth set/s since we're dealing with sets. These sets may be your PICKS to wage or a BASE to work on.

Lets go> Every lotto/lottery game has a POOL, with members ranging from lowest to highest, eg, cash 5 has lowest member 1 and highest 39(5/39>5 balls from pool of 39 balls), lottery games are usually 1 to highest 9(a ball from 9 balls). Let's set the variable N for the limit of the POOL. For any game , the format is x/N, where x is/are unknown variable/s to be drawn, for simplicity, i will assume the Linear dimension(yes I understand the dynamics of other dimensions), hence the variables will be drawn>Xo,X1........Xn. So for every draw set my Nth set will be >[(X1-X2),(X1-X2)+2 ,N-(X1-X2), N-(X1-X2)+2] . Note that X could be X3 ....Xn,

X1 and X2 assumes the variable X follows a linear sequence. The number 2 is deviation range for the variable X.

Now lets plug in some numbers using assumed cash 5 draw > 1-22-34-15-20:

United States Member #116344 September 8, 2011 3941 Posts Offline

Posted: November 9, 2013, 9:28 pm - IP Logged

Quote: Originally posted by adobea78 on November 9, 2013

This is a tentative formula derived from ' finding the next term' in a sequence. Whether such sequence is linear or not is left to observation. I use the word 'tentative' to avoid any equivocation . I will replace the Nth term, with Nth set/s since we're dealing with sets. These sets may be your PICKS to wage or a BASE to work on.

Lets go> Every lotto/lottery game has a POOL, with members ranging from lowest to highest, eg, cash 5 has lowest member 1 and highest 39(5/39>5 balls from pool of 39 balls), lottery games are usually 1 to highest 9(a ball from 9 balls). Let's set the variable N for the limit of the POOL. For any game , the format is x/N, where x is/are unknown variable/s to be drawn, for simplicity, i will assume the Linear dimension(yes I understand the dynamics of other dimensions), hence the variables will be drawn>Xo,X1........Xn. So for every draw set my Nth set will be >[(X1-X2),(X1-X2)+2 ,N-(X1-X2), N-(X1-X2)+2] . Note that X could be X3 ....Xn,

X1 and X2 assumes the variable X follows a linear sequence. The number 2 is deviation range for the variable X.

Now lets plug in some numbers using assumed cash 5 draw > 1-22-34-15-20:

United States Member #116344 September 8, 2011 3941 Posts Offline

Posted: November 9, 2013, 10:38 pm - IP Logged

Quote: Originally posted by adobea78 on November 9, 2013

Let s take one set at time and just filter via next draw,note your Remnant for your x

421 >2479-1380 fiter by 573> 24x9-1x80(Remnant 53,57

Picks 24x,29x,18x10x, where x=5>3>7

573>2479-4657 filter by 024>xx79-x657(remnant 0,02)Picks x79-65x-67x-57x

024>2479>filter923>x47(Remnant 3)picks x47

Remnant is simplist way of digit anaylsis without charts, is on going during workout, after

three workout my remnants are 5,3,5,7,0,0,2,3, and most of the time your elusive x is the

remnant

Since the assumption of the formula is linear in dimension, your picks should follow such dimension, this take care of positional picks most of the time. let's take above draw(see text) set 421 >2479-1380, convert it to a three set format>

United States Member #116344 September 8, 2011 3941 Posts Offline

Posted: November 11, 2013, 3:34 pm - IP Logged

Quote: Originally posted by adobea78 on November 9, 2013

This is a tentative formula derived from ' finding the next term' in a sequence. Whether such sequence is linear or not is left to observation. I use the word 'tentative' to avoid any equivocation . I will replace the Nth term, with Nth set/s since we're dealing with sets. These sets may be your PICKS to wage or a BASE to work on.

Lets go> Every lotto/lottery game has a POOL, with members ranging from lowest to highest, eg, cash 5 has lowest member 1 and highest 39(5/39>5 balls from pool of 39 balls), lottery games are usually 1 to highest 9(a ball from 9 balls). Let's set the variable N for the limit of the POOL. For any game , the format is x/N, where x is/are unknown variable/s to be drawn, for simplicity, i will assume the Linear dimension(yes I understand the dynamics of other dimensions), hence the variables will be drawn>Xo,X1........Xn. So for every draw set my Nth set will be >[(X1-X2),(X1-X2)+2 ,N-(X1-X2), N-(X1-X2)+2] . Note that X could be X3 ....Xn,

X1 and X2 assumes the variable X follows a linear sequence. The number 2 is deviation range for the variable X.

Now lets plug in some numbers using assumed cash 5 draw > 1-22-34-15-20:

This outcome gives constant 3/5 for lotto games and very effective for lottery games also.

Ideals are always welcome, keep your thoughts if you only find faults, am open to positive critique and ideals.

The best way to use the formula is to convert the results in $STRING, eg. if my draw set is 338 my end picks will be 0,2,9,1-5,7,4,6 >$string 02915746, now you have two options of filtering, you can start with draw set or the next set(remember to note the Remnants while filtering).

United States Member #116344 September 8, 2011 3941 Posts Offline

Posted: November 11, 2013, 4:00 pm - IP Logged

Quote: Originally posted by adobea78 on November 11, 2013

The best way to use the formula is to convert the results in $STRING, eg. if my draw set is 338 my end picks will be 0,2,9,1-5,7,4,6 >$string 02915746, now you have two options of filtering, you can start with draw set or the next set(remember to note the Remnants while filtering).

NB> Members of string are POSITIONAL.

Lets take CT data to illustrate the string workout using formula.

Pick 3

Pick 4

Midday

Evening

Midday

Evening

Mon, Nov 11, 2013

8-2-9

1-2-6-0

Sun, Nov 10, 2013

2-8-9

8-0-1

7-6-3-8

3-2-2-9

Sat, Nov 9, 2013

4-5-1

3-0-1

3-9-5-7

2-7-4-8

Fri, Nov 8, 2013

7-0-2

1-5-4

2-3-2-7

0-3-2-0

Thu, Nov 7, 2013

6-6-8

9-6-2

2-6-5-2

2-1-3-1

Wed, Nov 6, 2013

8-8-5

2-5-0

1-1-4-4

7-2-6-6

Tue, Nov 5, 2013

6-9-3

7-5-0

8-3-6-1

0-3-6-9

Mon, Nov 4, 2013

2-9-5

8-6-4

1-3-2-3

5-8-3-2

Sun, Nov 3, 2013

7-6-1

1-9-5

6-0-2-8

7-2-9-5

Sat, Nov 2, 2013

9-6-0

2-5-0

5-9-2-4

2-7-3-2

Start with P4, filtering will be combined draws>

2-7-3-2> 5746-4657-1380>filter by 6058>74-47-13(remnants 0,8,0,8,6,5)

74-47-13>filter by 7295>4-4-13(Remnants 2,9,5,2,9,5)

4-4-1-3(this is your base to wage.)

5-9-2-4>4657-7924-2479 >filter by 2732>465-94-49 (Remnants 2,3,3,3)

United States Member #133657 October 5, 2012 82 Posts Offline

Posted: November 11, 2013, 5:56 pm - IP Logged

When you say "This outcome gives constant 3/5 for lotto games" Do you mean just testing on paper or with actual numbers purchased and in play?

I've been doing a similar thing as you with generating smaller "play pools" with promising results on paper, but once the numbers are actually purchased and put into play the constant 3 numbers stop coming.

It's a curious phenomenon. I play, it stops working, I don't play, it starts working again. I'm just wondering if you have experienced this with your method? I look forward to reading more about it.

United States Member #116344 September 8, 2011 3941 Posts Offline

Posted: November 11, 2013, 7:28 pm - IP Logged

Quote: Originally posted by Lottonomics on November 11, 2013

When you say "This outcome gives constant 3/5 for lotto games" Do you mean just testing on paper or with actual numbers purchased and in play?

I've been doing a similar thing as you with generating smaller "play pools" with promising results on paper, but once the numbers are actually purchased and put into play the constant 3 numbers stop coming.

It's a curious phenomenon. I play, it stops working, I don't play, it starts working again. I'm just wondering if you have experienced this with your method? I look forward to reading more about it.

Backtesting can be biased. Most at times, we turn to work towards a known data, this creates inconsistency. I suggest you take any 10 draw sets and do a separate backtest without looking at actual results.

BETWEEN OAKRIDGE AND WRIGHT-PATTERSON AFB United States Member #1647 June 10, 2003 2729 Posts Offline

Posted: November 11, 2013, 7:54 pm - IP Logged

Quote: Originally posted by adobea78 on November 9, 2013

This is a tentative formula derived from ' finding the next term' in a sequence. Whether such sequence is linear or not is left to observation. I use the word 'tentative' to avoid any equivocation . I will replace the Nth term, with Nth set/s since we're dealing with sets. These sets may be your PICKS to wage or a BASE to work on.

Lets go> Every lotto/lottery game has a POOL, with members ranging from lowest to highest, eg, cash 5 has lowest member 1 and highest 39(5/39>5 balls from pool of 39 balls), lottery games are usually 1 to highest 9(a ball from 9 balls). Let's set the variable N for the limit of the POOL. For any game , the format is x/N, where x is/are unknown variable/s to be drawn, for simplicity, i will assume the Linear dimension(yes I understand the dynamics of other dimensions), hence the variables will be drawn>Xo,X1........Xn. So for every draw set my Nth set will be >[(X1-X2),(X1-X2)+2 ,N-(X1-X2), N-(X1-X2)+2] . Note that X could be X3 ....Xn,

X1 and X2 assumes the variable X follows a linear sequence. The number 2 is deviation range for the variable X.

Now lets plug in some numbers using assumed cash 5 draw > 1-22-34-15-20:

mid-Ohio United States Member #9 March 24, 2001 19900 Posts Offline

Posted: November 11, 2013, 8:09 pm - IP Logged

Quote: Originally posted by Lottonomics on November 11, 2013

When you say "This outcome gives constant 3/5 for lotto games" Do you mean just testing on paper or with actual numbers purchased and in play?

I've been doing a similar thing as you with generating smaller "play pools" with promising results on paper, but once the numbers are actually purchased and put into play the constant 3 numbers stop coming.

It's a curious phenomenon. I play, it stops working, I don't play, it starts working again. I'm just wondering if you have experienced this with your method? I look forward to reading more about it.

There no reason what you do on paper shouldn't work the same in reality. Next time when you're doing it on paper post the results on the prediction board too. If it works on the prediction board and on paper, chances are it will work in reality.

* you don't need to buy more tickets, just buy a winning ticket *